Tag Info

New answers tagged

1

Andromeda and the Milky Way belong to a group of galaxies called the Local Group. The two galaxies are the largest galaxies in the group, so to a pretty good approximation their interaction can be treated as a two body problem, with the other galaxies in the group producing only minor perturbations to their motion. So as you suspected, it isn't the case ...


0

Their mutual gravity will pull them towards each other, with the more massive galaxy causing more acceleration on the smaller. According to this article the more massive is the Milky Way, so it will cause Andromeda to accelerate more than the Milky Way, not that it really matters. Since interstellar space is mostly empty (i.e. there is a lot of distance ...


0

Strictly speaking, the universe has no rest frame (that we know of). If you want to set the rest frame of the CMB to be stationary, (which is reasonable in many applications), you may compare the velocity of the Milky Way and the velocity of Andromeda with respect to the CMB rest frame. (I don't know if anyone has ever done this, however.) The CMB dipole ...


0

In an inelastic collision, some of the energy is absorbed by the colliding bodies - this is why you cannot use conservation of energy to calculate the resultant velocities of the bodies involved - you don't know how much is absorbed. But you do know that momentum is conserved, and assuming that the bodies remain intact (no pieces are separated from the ...


0

Solving this particular set of equations would yield two values for $u1$ and as such two values for $u2$ - 10 and 5 ,that is, if the value of $u1$ were 5 the value of $u2$ would be 10 and vice-versa.You may arrive at this by solving a quadratic equation in $u1$ (or for that matter,$u2$) Now,you may have, while solving quadratics seen we,sometimes acquire ...


1

Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that ...


0

The answer is that there is no simple answer. The way that energy and momentum get split up in the aftermath of a collision depends on the details of the collision itself, and there is nothing in the conservation laws themselves that influences this. The simplest case is in one-dimensional collisions, where both objects are constrained to move along the ...


0

Isaac Newton observed the actions and reactions of objects in motion and recorded his observations as three famous laws. These laws (plus the conservation of momentum and of energy) can be used to explain how momentum and velocity are distributed among the objects coming out of a collision: (1) An object in motion will remain in motion with the same speed ...


0

The math is almost trivial for someone beyond algebra 1. Write the kinetic energy of each particle as $p_n^2/2m_n$. Then converse momentum and kinetic energy in the center-of-momentum. You will see that the magnitude of the momentum each particle does not change.


0

When the balls collide, they can transfer linear momentum. But the smooth surface means they transfer almost zero angular momentum. In the case where the cue ball takes a glancing shot, it transfers little momentum. There might be some spin on the ball, but the large amount of linear momentum means that it behaves almost exactly as you see in the ...


0

I also doubted it to begin with. Then I considered the case of A colliding with B directly on the A-B line. Then A stops cold and all its momentum is transferred to B. Now deviate the collision angle to the left by a very small amount, like 1 degree. What happens? A will almost stop. It will be left with a small velocity to the right, at right angles to ...


0

This actually makes sense if you look at the vector addition. If you add $\vec{v}_{2a}$ and $\vec{v}_{2b}$, you'll see they add up to $\vec{v}_{1a}$ for all elastic collisions with pool balls of equal mass. (Your vectors in your graphic are not to scale, so you can't do it graphically there.) This makes sense, because when the cue hits the other ball, it is ...


0

Ok, I might have "solved it" although there are still grey areas. First the equation for the force experienced by a body colliding against a body with infinite mass: $$ \textbf{F} = \textbf{n} \delta e_r m k_0 $$ Explanation: $ \textbf{n} $: unit normal pointing outside the colliding wall. This is the direction of the force (ie, where we want the ball ...


1

Given the mass m of the ball, the incident normal speed v, and the coefficient of restitution $\rho$, Then the integral of F over the duration of the collision $\Delta t$ is $$\int_0^{\Delta t} F dt = \frac{m(1 + \rho)v }{\Delta t}$$ assuning no rotational effects are incurred. This follows from the fact that at any instant the acceleration of the ball away ...


0

Simple momentum change will do: $$\vec F = \frac{d \vec p}{dt} $$ Your case is an elastic collision, so it simplifies to the difference between end states: $\vec F = \frac{\Delta \vec p}{\Delta t}$. Key thing: you need to know about the duration of the impact. Think about it... If your wall is very elastic - a vertical trampoline - the ball might still ...


1

I have found mistake in my calculatuion. First j = −(1 + e) vab1 · n j/= 1⁄ma + 1⁄mb + (rap × n)2 ⁄ Ia + (rbp × n)2 ⁄ Ib is how we calculate the impulse. taken from here http://myphysicslab.com/collision.html. the expression (rap × n) is a cross product of vectros, i calcualted it as dot product. and the expression jn is NOT a scalar value. it is ...


3

The second solution will mathematically satisfy the conservation equations, but corresponds the objects not actually colliding. Or they ``ghost'' and fly right through each other. :)


2

If you have a planet of mass $M$, then its self-gravitational binding energy is roughly $-GM^2/2R$ give or take a small numerical factor. So, for the Earth, this would be $-2\times 10^{32}$ J. Something colliding with the Earth, which has a similar mass and size, would do so at velocities of tens of km/s at least. I think the minimum closing velocity would ...


1

Anything over 500 miles in diameter, give or take is almost always sphere-shaped, the primary variation being rotation speed, which can give a flatness to the object, for example, Jupiter is visibly flattened by it's high rotational speed. The problem with building a strange shape by very large collision is that the heat generated in a collision of that ...


3

Q: When we write that, do we suppose a collisionless or collisional nature of the fluids? A: It's the energy-momentum tensor for a perfect fluid Chapter 2.26 Q: If this description corresponds to collisional fluids, why cosmological simulations are N-body simulations (collisionless) and are not simply based on hydrodynamics? A: Cosmological simulations are ...


1

I will yet add another formalism: Lets start with the hamiltonian form of Hamilton's Principle. Let $c: \mathbb R \longrightarrow T^*Q; t\mapsto (q(t),p(t))$ be the trajectory of a particle in the phase space of the configuration space $Q$, we define a subset of $Q$, $C$, where no contac occur between the particles, and $\partial C$ is te set of ponts wer ...


0

In the ballistic pendulum, it is the momentum that matters (that is conserved). For a pendulum with mass $M$, being hit by a bullet with mass $m$ and velocity $v$, the velocity $v_1$ immediately after the impact is (from conservation of momentum): $$v_1 = \frac{m}{m+M}v$$ And when you have an object with twice the mass but the same velocity, you get the ...


1

Depends on the nature of the collision. If there is a mechanism that takes energy from the system, i.e. a deformation, than energy is lost. You could think of your example as the center of mass of your rod as a point mass that starts rotating on a massless string once is passes the pivot. As usual, energy and momentum are conserved. You could do the ...



Top 50 recent answers are included